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Impulse and Momentum

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Linear Momentum“The change of motion is ever proportional to the motive force impressed; and is made in the direction of the right [straight] line in which that force is impressed”

Sir Isaac NewtonWhat Newton called “motion” translates into “moving inertia”.

!p = m!v

SI unit formomentum

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1 kilogram×1 meter

second=1

kg ⋅ms

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momentum = mass × velocity

momentum is avector quantity

Today the concept of moving inertia is called momentum which is defined as the product of mass and velocity.

LARGE MASS, SMALL VELOCITY

SMALL MASS, LARGE VELOCITY

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5/18/20

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Impulse and Momentum

Newton’s 2nd Law was written in terms of momentum and force

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F = ma

SI unit of impulse

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1 newton×1 second =1 N ⋅ s

impulse is avector quantity

Ft = mΔvimpulse momentum

change

Impulse causes a change of momentum for any object. This is analogous to work, which causes a change of energy for any object.

F = mΔvt

!I =!Ft 1 N ⋅ s = 1 kg ⋅m

s

F = Δpt

IMPULSE MOMENTUM THEOREM

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Impulse and Safety

MOMENTUM CHANGED BY A SMALL FORCE OVER A LONG TIME

MOMENTUM CHANGED BY A LARGE FORCE OVER A SHORT TIME

Other examples of car safety that involve increased time and decreased force (but result in equal impulse)

AirbagsSeatbeltsCrumple zonesBumpersPadding

Other examples force/time in impulse

Air cushioned shoesBending knees when landingNatural turf vs. artificial turfGym and playground matsHelmets/padding for sportsShocks/forks in bicycles

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Impulse of Sports

In many sports you are taught to “follow through”. Why?As you “follow through” the time of contact with the ball is increased, so the amount of momentum change is also increased. You get more momentum and more speed and more distance!

A boxer who “rolls with the punch” will experience less force over more time.

IN SPORTS THE IMPACT TIME IS

SHORT, BUT EVERY BIT COUNTS!

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BallMass(kg)

speed imparted

(m/s)

impacttime(ms)

Baseball 0.149 39 1.25

Football (punt) 0.415 28 8

Golf ball (drive) 0.047 69 1

Handball (serve) 0.061 23 12.5

Soccer ball (kick) 0.425 26 8

Tennis ball (server) 0.058 51 4

Impulse of Sports

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Third Law and ImpulsesEvery action has an equal and opposite reaction:

Every action takes just as long as the reaction so:

F1 = −F2

t1 = t2

The momentum changes are equal and opposite:m1Δv1 = −m2Δv2

F1t1 = −F2t2

Every impulse has an equal and opposite impulse:

The momentum changes are equal and opposite:Δp1 = −Δp2 or Δp1 + Δp2 = 0

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Conservation of Momentum

pi = pf or m1v1i + m2v2i = m1v1 f + m2v2 f

If two isolated objects interact (collide or separate), then the total momentum of the system is conserved (constant).

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ELASTIC COLLISIONINELASTIC COLLISIONSEPARATION OR

EXPLOSION

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Examples of Impulse and Momentum ProblemsIn 2007, Venus Williams hit the fastest recorded serve in a women’s match, at a speed of 58 m/s. What is the average force exerted on the 0.057-kg tennis ball by her racquet, assuming she hit the ball from rest, and the ball remained in contact with the racquet for 5.0 milliseconds.

In real-life collisions, the forces acting on an object are not constant. For example, when a bat strikes a baseball, the force increases with time, and then decreases, much like the figure to the right. However, if an effective force (or average) is know, the impulse-momentum theorem works! Suppose that a 42 m/s (94 mph) fastball is hit with an effective force of 760 N for 0.017 s. How fast does a 145 gm ball leave the bat if hit directly back at the pitcher?

An 8,000 kg railroad car is moving along a low-friction track at 3 m/s when a load of gravel is dropped into the railroad car. After the load is gravel is dropped, the railroad car moves at 2.5 m/s. How much is the mass of the dropped gravel?

Ft = mΔv ⇒ F(0.005 s) = (0.057 kg)(58 − 0 m/s) ⇒ F = 661 N

Ft = mΔv ⇒ (760 N)(0.017 s) = (0.145 kg)(vf − 42 m/s) ⇒ vf = 49.1 m/s or 110 mph!

m1v1i +m2v2i = m1v1 f +m2v2 f

(8000 kg)(3 m/s)+m2 (0) = (8000 kg +m2 )(2.5 m/s) ⇒ m2 = 1600 kg

Ft = mΔv

m1v1i +m2v2i = m1v1 f +m2v2 f

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Perfectly Inelastic CollisionsMost collision are somewhat inelastic, with some kinetic energy converted into thermal energy

m1v1i + m2v2i = (m1 + m2 )vf

When two objects collide they may combine into one object in a perfectly inelastic collision

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Elastic CollisionsIdeal collisions are elastic, with zero kinetic energy converted into thermal energy (heat)Springs and magnets can create elastic collisions. Other examples: air molecules, billiard balls, spacecraft gravitational boost.

NEWTON’S CRADLE

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Examples of Inelastic and Elastic CollisionsTrain cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 1.50×105 kg and a velocity of 0.30 m/s to the right, and the second having a mass of 1.10×105 kg and a velocity of 0.15 m/s to the left, as shown in the figure. What is their final velocity?

A 6.3-kg bowling ball moving at 9.0 m/s collides directly with a 1.5-kg bowling pin, which is sent forward with a speed of 13.0 m/s. Calculate the final velocity of the bowling ball. Is the collision elastic?

HONORS: In an elastic collision, a 450-kg bumper car collides directly from behind with a second, 400-kg car that is traveling in the same direction. The initial speed of the leading bumper car is 4.8 m/s and that of the trailing car is 6.0 m/s. What are their final speeds? (Assume the masses include the drivers.)

m1v1i +m2v2i = (m1 +m2 )vf

m1v1i +m2v2i = m1v1 f +m2v2 f

m1v1i +m2v2i = m1v1 f +m2v2 f and 12 m1v1i

2 + 12 m2v2i

2 = 12 m1v1 f

2 + 12 m2v2 f

2 combine to give:

KEi ≠ KEf ?? ⇒ 12 m1v1i

2 ≠ 12 m1v1 f

2 + 12 m2v2 f

2 ⇒ 12 (6.3)(92 ) ≠ 1

2 (6.3)(5.92 )+ 12 (1.5)(132 ) ⇒ 255 J > 236 J

450(6)+ 400(4.8) = 450v1 f + 400v2 f and 6 − 4.8 = −(v1 f − v2 f ) ⇒ v2 f = v1 f +1.2

4620 = 450v1 f + 400(1.2 + v1 f ) ⇒ v1 f = (4620 − 400(1.2)) / 850 = 4.87 ms , and v2 f = 6.07 m

s

(1.5 ×105 )(0.3)+ (1.1×105 )(−0.15) = (1.5 ×105 +1.1×105 )vf ⇒ vf = 0.11m/s

6.0 m/s 4.8 m/s

⇒ (6.3)(9)+ (1.5)(0) = (6.3)v1 f + (1.5)(13) ⇒ v1 f = 5.90 m/s

v1i − v2i = −(v1 f − v2 f ) which is a "relative velocity" equation. In other words, vrel ,approach = -vrel ,recede

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Examples of Momentum and Energy ProblemsTwo carts are placed on a frictionless track that has a spring of constant k = 45 N/m attached to one end, as shown. m1 = 0.75 kg has aninitial velocity 3.5 m/s to the right and m2 = 0.50 kgis initially at rest. (a) If the carts collide and stick,find the velocity of the carts just after the collision.(b) Find the maximum compression in the spring.

HONORS: A pellet of mass m is fired into a block of mass M initially at rest at the edge of a frictionless table of height h as shown. The pellet remains in the block, and after impact the block lands a distance d from the bottom of the table. Calculate the initial speed of the pellet in terms of m, M, h, d and g.

A block of mass m starts from rest and slides down a frictionless track from a height h = 0.9 m as shown. When it reaches the lowest point of the track, it collides with a stationary piece of putty of mass 2m. The block and the putty stick together and continue to slide. What is the maximum height that the block-putty can reach?

m1v1i +m2v2i = (m1 +m2 )vf

m1v1i +m2v2i = m1v1 f +m2v2 fΔx = vxt ⇒ d = vf t and Δy = 1

2 gt2 ⇒ h = 0t + 1

2 gt2 ⇒ t = 2h / g

⇒ (0.75)(3.5)+ (0.50)(0) = (0.75 + 0.50)vf ⇒ vf = 2.10 m/s

⇒ mv1i +M (0) = (m +M )vf

3.5 m/s

m1 m2

k

⇒ 12 (1.25)(2.12 ) = 1

2 (45)(x2 ) ⇒ x = 0.35 mKEi = EPEf ⇒ 12 mv

2 = 12 kx

2

GPEi = KEf ⇒ mgh1 = 12 mv2

2

mv2 + 0 = (m + 2m)v3

m

2m

h?

at rest

at rest

⇒ 12 3mv3

2 = 3mgh4 ⇒ h4 =v3

2

2g= 1.42

2(9.8)= 0.1 m

⇒ v2 = 2gh1 = 2(9.8)(0.9) = 4.2 ms

⇒ v3 =m

m + 2m(4.2) = 1.4 m

s

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2 3

4

vf = d / t = d / 2h / g and v1i =(m +M )

mvf =

(m +M )dm

g2h

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Ballistic Pendulum ExampleA useful device for measuring the speed of a ballistic (in the 18th century) combines the use of conservation of energy and conservation of momentum.A bullet is fired into a heavy block of wood, which swings upward with the bullet embedded inside. The height of the swinging pendulum is measured and the bullet’s speed is then calculated.

A bullet of mass 45.0 g is fired into a wooden block of mass 5.3 kg. The block–bullet pendulum rises to a maximum height of 13.0 cm above the block's initial height.

Which conservation law is useful for the bullet/block collision? Which is best for the pendulum swing? Why?

(0.045)(v1i )= (0.045+5.3)(1.60)⇒ v1i =190 m/s

mbulletv1i +mblockv2i = (mbullet +mblock )v

v = 2gh = 2(9.8 N/kg)(0.13 m) =1.60 m/s

12 mbullet +mblock( )v2 = mbullet +mblock( )gh

HONORS: what angle does the pendulum swing to if L = 40 cm?

θ = cos−1(1− h

L )= cos−1(1− 13

40 )= 47.5˚

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